Standard summation results

There are a set of results of \sum which you need to learn.

They are:

• \sum^n_{r=1} 1
• \sum^n_{r=1} r
• \sum^n_{r=1} r^2
• \sum^n_{r=1} r^3

\sum^n_{r=1} 1

• This means that we are adding 1 a total of n times.
• So the result is n.

\sum^n_{r=1} 1 = n

\sum^n_{r=1} r

• This means that we are adding all the integers from 1 to n.
• The result is the triangular number n!
• The formula for finding triangle numbers is \frac{n(n+1)}{2}.

\sum^n_{r=1} r = \frac{n(n+1)}{2}

\sum^n_{r=1} r^2

• This means that we are adding the squares of all the integers from 1 to n.
• The formula for finding this is \frac{n(n+1)(2n+1)}{6}.

\sum^n_{r=1} r^2 = \frac{n(n+1)(2n+1)}{6}

\sum^n_{r=1} r^3

• This means that we are adding the cubes of all the integers from 1 to n.
• The formula for finding this is \left(\frac{n(n+1)}{2}\right)^2.

\sum^n_{r=1} r^3 = \left(\frac{n(n+1)}{2}\right)^2

Deducing results

Using standard results, deduce the value of \sum^n_r=1(2r-3)

• We can rewrite the above as \sum^n_{r=1} 2r - \sum^n_{r=1} 3
• Which we can use the distributive law on:
  • 2\sum^n_{r=1} r - 3\sum^n_{r=1} 1
• We can substitute in our standard results:
  • 2(\frac{n(n+1)}{2}) - 3(n)
• Which simplifies to:
  • n(n+1) - 3n
  • n^2 + n - 3n
  • n^2 - 2n
• Answer: \sum^n_{r=1} (2r-3) = n^2 - 2n

Using standard results, deduce the value of \sum^{23}_{18} r^2

• The values in the range of 18-23 will add up to the same as the values from 1-23 minus the values from 1-17:
  • \sum^{23}_{r=18} r^2 = \sum^{23}_{r=1} r^2 - \sum^{17}_{r=1} r^2
• We can substitute in our standard results:
  • \frac{23(23+1)(2\times23+1)}{6} - \frac{17(17+1)(2\times17+1)}6
• Which simplifies to:
  • \frac{23\times24\times47}{6} - \frac{17\times18\times35}{6}
  • 4324 - 1785
  • 2539
• Answer: \sum^{23}_{r=18} r^2 = 2539

Using standard results, deduce the value of \sum^8_1(2r^3-3r^2)

• =\sum^8_1(2r^3)-\sum^8_1(3r^2)
• =2\sum^8_1(r^3)-3\sum^8_1(r^2)
• TODO

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